Which limit is correct? Why does L'Hospital's rule yield a different result? [closed]

https://i.stack.imgur.com/cIf8R.jpg

I would appreciate it if someone could explain why the two methods yield different results to the same limit. Have I done something incorrect? I can't seem to find the error in either method. And it would be much appreciated if the correct answer was pointed out as well.

• Line number $4$ of the first part is incorrect : how can you replace $\frac{\ln(1+x)}{x^2}$ by $\frac{1}{x}$ ? Aug 27, 2020 at 14:39
• I used the standard limit for the natural logarithm of (1+x) all over x as x tends to 0. Since this limit is 1, one of the x in the denominator is left over. Is that not applicable here? Aug 27, 2020 at 14:46
• When you have a sum, you have to be careful with the interactions. See my answer. Aug 27, 2020 at 14:48
• $\ln(1+x)/x$ tends to $1$, but is not equal to $1$. So you are not allowed to replace it by $1$. Aug 27, 2020 at 14:48
• In general, try to type out your question. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Aug 27, 2020 at 15:35

On the left side, you cannot replace $$\ln(1+x)/x^2$$ by $$1/x$$, because it is not a factor in your expression, but an addend. You would have to make sure that you are not missing terms that could contribute to the leading term. In your case, you would have to expand to the next term: $$\ln(1+x)/x^2=1/x-1/2+\dots$$. The $$1/x$$ cancels but the constant $$-1/2$$ contributes to the limit, since it is of the same order as $$\ln(1+x)/x=1+\dots$$. The final result is $$1/2$$, as on the right side. Hope this helps.

• So just to clarify, you can't just evaluate limits separately if it's an addend. If there is a common factor, then it's allowed. But that brings me to wonder....why do we say that limits are additive? Like the limit of f(x) + g(x) = limit f(x) + limit g(x)? Aug 27, 2020 at 14:54
• (+1) This is a clear and concisely written solution. Aug 27, 2020 at 15:51
• @sider003 Because $\ln(1+x)/x^2$ doesn't have a limit. Aug 27, 2020 at 16:02

To simplify without error, you have to expand the numerator at order $$2$$ atleast, since the denomunatot is $$x^2$$, and all terms have to be expanded at the same order. What you did is that you replaced $$\frac{\ln(1+x)}{x^2}$$ with its equivalent $$\frac x{x^2}$$, thereby forgetting that equivalence of function is not compatible with addition or subtraction.

Here is how I would have done it:

• $$(1+x)\ln(1+x)=(1+x)\Bigl(x-\frac{x^2}2+o(x^2)\Bigr)\\=x+x^2+o(x^2)-\smash{\underbrace{\frac{x^2}2-\frac{x^3}2++xo(x^2)}_{o(x^2)}}\\[1ex]=x+\frac{x^2}2+o(x^2)$$
• Therefore, the fraction is $$\frac{x+\frac{x^2}2+o(x^2)-x}{x^2}=\frac{\frac{x^2}2+o(x^2)}{x^2}=\frac12+o(1).$$

$$\require{cancel}$$ \begin{align} & \lim_{x\to0} \left( \frac{\log(1+x)}{x^2} - \frac 1 x \right) \\[10pt] = {} & \lim_{x\to0} \left( \left(\frac{\log(1+x)} x \right) \cdot \frac 1 x - \frac 1 x \right) \\[12pt] = {} & \xcancel {\left( \lim_{x\to0} \frac{\log(1+x)} x \right)\bullet \left( \lim_{x\to0} \frac 1 x \right) - \left( \lim_{x\to0} \frac 1 x \right)} \quad \text{?} \\ & \text{This does not work because those} \\ & \text{last two limits are not finite.} \\[15pt] = {} & \xcancel {\left( \lim_{x\to0} \frac{\log(1+x)} x \right)\bullet \left( \lim_{x\to0} \left( \frac 1 x - \frac 1 x \right) \right)} \quad \text{?} \\ & \text{This does not work because } \frac 1 x - \frac 1 x \text{ is not} \\ & \text{what that first fraction what multiplied by.} \end{align}